The strong interaction cannot tell a proton from a neutron. Everything in these three sections is a consequence of that blindness, including the parts that look like predictions about cross-sections.
🎯 Why this matters
An approximate symmetry is more useful here than an exact one would be. Electromagnetism breaks isospin and the strong force does not, so a violation is a label: it says which interaction was responsible.Every quantum number so far has been either multiplied (P, C) or counted (B, L, S). Isospin isospin I the SU(2) quantum number under which proton and neutron are two states of one particle; a multiplet has 2I + 1 members sharing mass, spin and parity and differing only in I_z. Conserved by the strong interaction only. defined in §3.1-3.2 — open in glossary is the first one that is rotated — a continuous symmetry acting in a space that is not physical space at all — and it is the most productive idea in the chapter. It sorts the hadrons into multiplets, it explains why their masses come in near-degenerate clusters, and in §3.9 it predicts ratios of cross-sections with no dynamical input whatsoever. That last result, 9 : 1 : 2 against a measured 195 : 22 : 45, is the best single number in this chapter.
3.8 Isospin
The nuclear force barely notices electric charge. The binding energies of ³H (1 proton, 2 neutrons) and ³He (2 protons, 1 neutron) are very nearly equal, and what difference there is comes from the electrostatic repulsion between the two protons in ³He — an electromagnetic effect, not a nuclear one.
Heisenberg’s 1932 proposal was stronger than “the force is blind to charge”. It was that the proton and the neutron are two states of one particle, the nucleon, and that the nuclear force is invariant under rotations in the internal space that mixes them.
📐 Physics you need first — an internal space that is not space
Every symmetry you have met so far acts on something you can point at. Parity inverts the coordinate axes; a Lorentz boost changes a velocity; a rotation turns a laboratory. Isospin does none of that. It acts on a two-dimensional complex vector space whose basis vectors are labelled “proton” and “neutron”, and rotating in it turns one into the other without moving anything.
That is a genuinely new kind of object, so it is worth being precise about what carries over from ordinary angular momentum and what does not.
What carries over — all the algebra. The generators obey the same commutation relations, so every result about adding angular momenta applies unchanged: an isospin multiplet isospin multiplet the set of hadrons or nuclear levels related by isospin rotations; degenerate but for the electromagnetic interaction and the u–d mass difference, which split them by a few MeV. defined in §3.8-3.10 — open in glossary has members, runs in integer steps from to , and two isospins combine with the very same Clebsch–Gordan coefficients tabulated in Appendix 4. This is why the machinery of §3.9 needs no new mathematics at all.
What does not carry over — the meaning. is not a projection along any laboratory axis, so “rotating your apparatus” does nothing to it, and there is no isospin analogue of an angular-momentum measurement in a Stern–Gerlach magnet. What labels is which member of the multiplet you have — which is to say, the electric charge, through Eq. (3.39) below.
And it is only approximate. Rotation invariance in real space is exact. Isospin invariance is not: the electromagnetic interaction distinguishes the proton from the neutron because it couples to charge, and the u and d quarks do not have quite the same mass. Both effects are small — a few MeV — which is why the multiplets are nearly, and not exactly, degenerate.
📏 Sign convention — and a slip on p. 119
The book fixes the convention explicitly: the nucleon has , with the proton at and the neutron at . Page 119 prints for both, which is a typo — the two members of a doublet cannot share a third component, and the Gell-Mann–Nishijima relation Gell-Mann–Nishijima relation the exact identity Q = I_z + Y/2 relating a hadron's electric charge to its isospin and hypercharge; an assertion you can run over the whole particle table and expect zero failures. defined in §3.8-3.10 — open in glossary (3.39) gives the neutron immediately.
It matters because every sign in §3.9 descends from it. For a nucleus of charge and mass number the same convention gives , so the triplet below runs , , . The text on p. 120 lists the three values in the opposite order; nothing downstream depends on it, but the reader checking the arithmetic should know which way round it goes.
A multiplet you can see: the A = 12 triplet
- the J^P = 1⁺, I = 1 triplet
- ¹²C ground state, I = 0
💡 What this really says — three states in three nuclei are one object in three charge states
Three states, belonging to three different nuclei, share a spin and a parity () and sit at almost the same energy. Isospin says they are not three things: they are one object, seen in its three charge states, and the small energy differences are a perturbation on top.
Two sources produce the splitting, and both are outside the strong interaction:
- the electromagnetic interaction, which does not conserve (it does conserve — see the table in §3.1), and which costs a nucleus more the more protons it has to hold together;
- the quark masses, since exceeds by a few MeV, which is also why MeV in the first place.
Turn both off and the three levels coincide exactly.
⚙️ Engineer’s bridge — a multiplet is a degenerate eigenspace
You have solved this problem, in a different vocabulary.
A symmetric system has degenerate modes: two identical pendulums coupled symmetrically, a square membrane, a balanced three-phase network. Degeneracy is never an accident — it is forced by a symmetry of the operator, and the degenerate states span an invariant subspace that the symmetry rotates among themselves. That subspace is the multiplet, and its dimension is the dimension of an irreducible representation.
Then break the symmetry slightly and the degeneracy lifts. Add a small asymmetry to the coupled pendulums and the single frequency splits into two, separated by an amount proportional to the perturbation. That is precisely the 3 MeV spread above, and precisely the 1.3 MeV between the neutron and the proton.
The engineering payoff is the diagnostic reading: a cluster of nearly degenerate levels is evidence of an approximate symmetry, and the size of the splitting measures how badly it is broken. Physicists found isospin the same way you would find a hidden symmetry in a system you had not modelled — by noticing that the spectrum came in suspiciously even groups.
Where the analogy breaks: coupled pendulums live in a real vector space and the symmetry is a rotation you could perform. Isospin’s SU(2) acts on a complex two-dimensional space, and no experiment can perform the rotation — you can only observe that the physics is invariant under it.
Where it breaks: the degenerate-eigenspace picture suggests you could rotate within the multiplet and check the physics is unchanged, the way you would test a symmetry in software. You cannot: no apparatus performs an isospin rotation. A proton is not turned into a neutron by any knob, so the symmetry is verified only through its consequences — equal masses, related cross-sections, forbidden channels — never by doing the transformation. A symmetry you can only observe the shadow of is a weaker epistemic position than one you can apply.
SU(2), and the two quantum numbers that label a hadron
The book switches from three-dimensional rotations to SU(2) — the group of 2 × 2 unitary matrices of unit determinant. The two are equivalent for this purpose, and the reason for the switch is forward-looking: SU(2) generalises to SU(3) in Chapter 4, while “rotations in three dimensions” does not.
| dimension | 1 | 2 | 3 | 4 | 5 | … |
|---|---|---|---|---|---|---|
| isospin | 0 | 1/2 | 1 | 3/2 | 2 | … |
Isospin alone is not enough to organise the hadrons, so §3.9 adds a second additive quantum number, the hypercharge hypercharge Y Y = B + S, constant across an isospin multiplet and conserved wherever strangeness is. For mesons it is simply the strangeness. defined in §3.8-3.10 — open in glossary .
That is Table 3.2, and it is the whole of the representation theory needed here: the number of members of a multiplet is the label. Leptons and the photon do not appear in it at all — they have no isospin, because isospin is a property of the strong interaction and they do not feel it.
Bettini p. 120 — the flavour hypercharge. Constant across a multiplet, which is what makes it a useful second axis.
Every symbol, one at a time
Hover or tap a symbol above — it lights up in the equation and its meaning, units and type appear here.
Bettini p. 121 — the Gell-Mann–Nishijima relation. It converts a quantum number nobody can measure into three that everybody can.
Every symbol, one at a time
Hover or tap a symbol above — it lights up in the equation and its meaning, units and type appear here.
💡 What this really says — why EM conserves I_z but not I
The right-hand side of (3.39) is built entirely from , and , and the electromagnetic interaction conserves all three separately. So it must conserve — as an arithmetic consequence, with no further physics.
But it does not conserve , because it distinguishes the members of a multiplet by their charge, and a rotation in isospin space is exactly what mixes them. The result is the odd-looking row in the symmetry table: violated, conserved, by the same interaction.
The clean example is , which takes with throughout. It happens, it is electromagnetic, and it changes — which is the whole distinction in one decay.
The multiplets themselves
Fig. 3.2 — the J^P = 1/2⁺ baryons · JP = 1/2⁺
Click any member for its quantum numbers — and for the Gell-Mann–Nishijima check Iz = Q − Y/2.
Masses are the book's rounded figure values in MeV. Note that the Λ⁰ and the Σ⁰ share a hypercharge and a quark content and are still different particles — they differ in isospin, which is what a multiplet label is for. Press the antiparticle button for Question 3.3: every charge flips, so the whole diagram reflects through the origin.
Fig. 3.3 — the J^P = 0⁻ mesons · JP = 0⁻
Click any member for its quantum numbers — and for the Gell-Mann–Nishijima check Iz = Q − Y/2.
For mesons B = 0, so Y is simply the strangeness. The π⁻ and π⁺ are each other's antiparticles inside ONE multiplet, and the π⁰ is its own — but the kaons need two separate doublets, one for the particles and one for the antiparticles, because a K⁺ and a K⁻ differ in strangeness by two units and cannot be members of the same I = 1/2 row.
🔢 Worked example — Gell-Mann–Nishijima on every member
Equation (3.39) is a constraint, not a definition of convenience, so it is worth checking on all fifteen hadrons above rather than on one.
Reproduce it
H = [ # label, Q, B, S, Iz — read off Figs. 3.2 and 3.3
("n", 0, 1, 0, -0.5), ("p", +1, 1, 0, +0.5),
("Lambda0", 0, 1, -1, 0.0),
("Sigma-", -1, 1, -1, -1.0), ("Sigma0", 0, 1, -1, 0.0), ("Sigma+", +1, 1, -1, +1.0),
("Xi-", -1, 1, -2, -0.5), ("Xi0", 0, 1, -2, +0.5),
("K0", 0, 0, +1, -0.5), ("K+", +1, 0, +1, +0.5),
("pi-", -1, 0, 0, -1.0), ("pi0", 0, 0, 0, 0.0), ("pi+", +1, 0, 0, +1.0),
("K-", -1, 0, -1, -0.5), ("K0bar", 0, 0, -1, +0.5),
]
print(f"{'hadron':9s}{'Q':>3s} {'B':>4s} {'S':>4s} {'Y=B+S':>8s} {'Q-Y/2':>7s} {'Iz':>5s} ok")
ok = True
for lab, Q, B, Sq, Iz in H:
Y = B + Sq
calc = Q - Y / 2
good = abs(calc - Iz) < 1e-12
ok &= good
print(f"{lab:9s}{Q:+3d} {B:4d} {Sq:4d} {Y:+8d} {calc:+7.1f} {Iz:+6.1f} {'yes' if good else 'NO'}")
print(f"all {len(H)} hadrons satisfy Iz = Q - Y/2" if ok else "MISMATCH")
tot = {}
for lab, Q, B, Sq, Iz in H:
tot[(B, Sq if B else 0, B + Sq)] = tot.get((B, Sq if B else 0, B + Sq), 0) + Iz
print("multiplet sums: every multiplet has sum(Iz) = 0, as a multiplet must"
if all(abs(v) < 1e-12 for v in tot.values()) else "a multiplet does not sum to zero")hadron Q B S Y=B+S Q-Y/2 Iz ok n +0 1 0 +1 -0.5 -0.5 yes p +1 1 0 +1 +0.5 +0.5 yes Lambda0 +0 1 -1 +0 +0.0 +0.0 yes Sigma- -1 1 -1 +0 -1.0 -1.0 yes Sigma0 +0 1 -1 +0 +0.0 +0.0 yes Sigma+ +1 1 -1 +0 +1.0 +1.0 yes Xi- -1 1 -2 -1 -0.5 -0.5 yes Xi0 +0 1 -2 -1 +0.5 +0.5 yes K0 +0 0 1 +1 -0.5 -0.5 yes K+ +1 0 1 +1 +0.5 +0.5 yes pi- -1 0 0 +0 -1.0 -1.0 yes pi0 +0 0 0 +0 +0.0 +0.0 yes pi+ +1 0 0 +0 +1.0 +1.0 yes K- -1 0 -1 -1 -0.5 -0.5 yes K0bar +0 0 -1 -1 +0.5 +0.5 yes all 15 hadrons satisfy Iz = Q - Y/2 multiplet sums: every multiplet has sum(Iz) = 0, as a multiplet must
Fifteen hadrons, fifteen agreements, and no free parameter anywhere. The relation also explains a fact the figures make visually obvious: the multiplet is centred on , so the average charge of a multiplet is . The nucleon doublet averages ; the pion triplet averages 0; the Ξ doublet averages .
3.9 Summing isospins, and the product of representations
Isospin earns its place here. Classifying hadrons is useful; constraining their dynamics without knowing the dynamics is remarkable.
The rules follow from the symmetry table of §3.1:
- strong interactions conserve both and ;
- electromagnetic interactions conserve only ;
- weak interactions conserve neither.
So a strong process must have at least one value of total isospin common to its initial and final states, and each such value gets its own isospin amplitude isospin amplitude one complex amplitude per total-isospin channel (A_{1/2}, A_{3/2}); every cross-section in a related family is a Clebsch–Gordan combination of the same few, so ratios follow with no dynamical input at all. defined in §3.8-3.10 — open in glossary — a complex, unknown, uncomputable function of the kinematics. Every reaction in the family is a different linear combination of the same amplitudes, which is why the ratios are predicted and the absolute values are not.
Bettini pp. 122–123. The same statement twice: on the left labelled by isospin, on the right by the number of states — the notation that survives into SU(3).
Every symbol, one at a time
Hover or tap a symbol above — it lights up in the equation and its meaning, units and type appear here.
⚙️ Engineer’s bridge — this is block-diagonalisation, and Clebsch–Gordan is the change of basis
product of representations the decomposition of a product of SU(2) representations into a sum, 1 ⊗ ½ = ½ ⊕ 3/2, written in dimensions as 3 ⊗ 2 = 2 ⊕ 4; the notation that carries over to SU(3) in ch04. defined in §3.8-3.10 — open in glossary is a statement about matrices, and reading it that way makes the whole of §3.9 mechanical.
The πN system has six states, so any isospin rotation acts on it as a 6 × 6 unitary matrix. In the product basis — “which pion, which nucleon” — that matrix is dense and tells you nothing. Change to the total-isospin basis and the same matrix becomes block diagonal: one 4 × 4 block and one 2 × 2 block, which never mix. is the dimension check, and it is Eq. (3.40) in words.
- The Clebsch–Gordan coefficients are the entries of that change-of-basis matrix. The Appendix 4 calculator already shows them both ways for exactly this reason — as a table of coefficients, and as a unitary matrix whose blocks you can see.
- A conserved quantity is a block label. Because the strong interaction commutes with isospin rotations, its matrix is block diagonal in the same basis — so it carries one number per block, and those numbers are and .
- Selection rules are the zeros. Off-block entries vanish identically, which is the sparsity pattern that forbids a transition.
The engineering reading: you have diagonalised an operator using a symmetry rather than by computing eigenvalues, and the payoff is that a completely unknown operator is now described by two numbers instead of a 6 × 6 matrix. The ratios below are what is left when two unknowns are shared among six measurements.
Where it breaks: block-diagonalisation compresses six measurements into two unknowns only if the states really are pure isospin eigenstates, and they are not exactly. Electromagnetic and quark-mass effects mix the blocks at the per-cent level, so the predicted ratios carry an irreducible ~1 % error that no better measurement removes — and near a resonance where two isospin amplitudes have different phases the interference term makes the discrepancy larger than that. The compression is excellent and it is lossy.
Dimension check: 6 = 4 + 2 = 6 ✓ · rows orthonormal ✓ — the table is an orthogonal matrix, so reading it backwards is just its transpose.
Is this reaction allowed? — Examples 3.3 to 3.6
| reaction | initial I | final I | verdict | |
|---|---|---|---|---|
| π⁻ p → π⁰ n | 1 ⊗ ½ = ½ ⊕ 3/2 | 1 ⊗ ½ = ½ ⊕ 3/2 | −½ → −½ | |
| d d → ⁴He π⁰ | 0 ⊗ 0 = 0 | 0 ⊗ 1 = 1 | 0 → 0 | |
| π⁰ → γγ | 1 | 0 (the photon has no isospin) | 0 → 0 | |
| Λ → p π⁻ | 0 | ½ ⊗ 1 = ½ ⊕ 3/2 | 0 → −½ |
The method is always the same three steps: decompose both sides into total isospin, look for a common value, then check I_z separately. <strong>Which of the three tests fails tells you which interaction is responsible</strong> — and that is a conclusion about dynamics reached without any dynamics.
The four πN reactions, and 9 : 1 : 2
Bettini p. 124. The change of basis, written out — and the object in the middle is the Clebsch–Gordan coefficient.
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Bettini p. 124. The two extreme states are pure I = 3/2 — there is simply nothing else with |I_z| = 3/2 — while the two middle ones are mixtures.
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💡 What this really says — the whole result follows from which states are pure
Look at which states are pure and which are mixed, because the whole result follows from that asymmetry.
has . The only total isospin that can accommodate is , so the state is pure — no decomposition, no coefficient, no ambiguity. The same holds for at .
and both have , where both total isospins are available. They are therefore mixtures of the same two kets — with different coefficients, and, crucially, with a relative minus sign in one of them. That sign is what makes the elastic and charge-exchange channels carry different combinations, and therefore what makes the family solvable.
So: two pure reactions measure on their own, and the mixed ones then constrain and the relative phase. Six measurements, two unknowns.
Bettini p. 124. One constant K, common to all four, and two complex amplitudes. Everything else is Clebsch–Gordan arithmetic.
Every symbol, one at a time
Hover or tap a symbol above — it lights up in the equation and its meaning, units and type appear here.
💡 What this really says — one prediction that needs no amplitudes at all
Before any resonance is invoked, (3.42) and (3.45) already give a result:
at every energy, because both reactions are pure and therefore carry the identical amplitude. No knowledge of is needed — it cancels between the two sides. The book notes it is experimentally well verified, and it is the cleanest possible test of isospin symmetry itself: an equality between two measurements, with no free parameter and no model.
Set the sliders in the widget below to anything you like and watch those two rows stay locked together.
πN cross-section ratios from isospin alone
| reaction | amplitude | σ (arb.) | ÷ σ(π⁻p elastic) |
|---|---|---|---|
| π⁺p → π⁺p(3.42) | 1 A₃⁄₂ | 1.0000 | 9.00 |
| π⁻p → π⁻p(3.44) | 1/3 A₃⁄₂ + 2/3 A₁⁄₂ | 0.1111 | 1.00 |
| π⁻p → π⁰n(3.43) | √2/3 A₃⁄₂ − √2/3 A₁⁄₂ | 0.2222 | 2.00 |
| π⁻n → π⁻n(3.45) | 1 A₃⁄₂ | 1.0000 | 9.00 |
| π⁺n → π⁺n | 1/3 A₃⁄₂ + 2/3 A₁⁄₂ | 0.1111 | 1.00 |
| π⁺n → π⁰p | √2/3 A₃⁄₂ − √2/3 A₁⁄₂ | 0.2222 | 2.00 |
This is Eq. (3.46). With the I = 1/2 amplitude switched off, the three cross-sections stand in the ratio 9 : 1 : 2, and the measured 195 : 22 : 45 gives 21.7 and 43.3 against 22 and 45. Nothing about the strong interaction was used — only how the states decompose.
🔢 Worked example — Eq. (3.46), and how well 9 : 1 : 2 does
At low energy all these cross-sections show a large resonance, the , discovered by Fermi’s group in 1952. Its isospin is , which shows up as — so set and read off (3.42), (3.44) and (3.43):
The measured values in millibarns are 195 : 22 : 45.
Reproduce it
from math import sqrt, factorial as f
def cg(j1, m1, j2, m2, J, M): # Racah, Condon-Shortley (= src/lib/cg.ts)
if m1 + m2 != M or abs(M) > J or J > j1 + j2 or J < abs(j1 - j2): return 0.0
pre = sqrt((2*J+1)*f(int(j1+j2-J))*f(int(j1-j2+J))*f(int(-j1+j2+J))/f(int(j1+j2+J+1)))
pre *= sqrt(f(int(j1+m1))*f(int(j1-m1))*f(int(j2+m2))*f(int(j2-m2))*f(int(J+M))*f(int(J-M)))
s = 0.0
for k in range(40):
d = [j1+j2-J-k, j1-m1-k, j2+m2-k, J-j2+m1+k, J-j1-m2+k]
if any(x < 0 or x != int(x) for x in d): continue
s += (-1)**k / (f(k)*f(int(d[0]))*f(int(d[1]))*f(int(d[2]))*f(int(d[3]))*f(int(d[4])))
return pre * s
st = {'pi+ p': (1, .5), 'pi- p': (-1, .5), 'pi0 n': (0, -.5), 'pi- n': (-1, -.5)}
print("decomposition from lib/cg.ts (Condon-Shortley), |pi N> in the |I,Iz> basis")
for k, (m1, m2) in st.items():
parts = "".join(f"{cg(1,m1,.5,m2,J,m1+m2):+8.4f} |{'3/2' if J==1.5 else '1/2'},"
f"{'+' if m1+m2>=0 else '-'}{abs(int(2*(m1+m2)))}/2> "
for J in (1.5, .5) if abs(cg(1,m1,.5,m2,J,m1+m2)) > 1e-9)
print(f" |{k:6s}> = {parts}")
amp = lambda a, b, A32, A12: sum(cg(1,st[a][0],.5,st[a][1],J,sum(st[a]))
* cg(1,st[b][0],.5,st[b][1],J,sum(st[b])) * A
for J, A in ((1.5, A32), (.5, A12)))
print("cross-sections with A_1/2 = 0 (the Delta dominates):")
lab = [('pi+ p','pi+ p','3.42'), ('pi- p','pi- p','3.44'),
('pi- p','pi0 n','3.43'), ('pi- n','pi- n','3.45')]
s = []
for a, b, eq in lab:
v = amp(a, b, 1.0, 0.0)**2; s.append(v)
print(f" {a.replace(' ','')} -> {b.replace(' ','')} |amp|^2 = {v:.5f} ({eq})")
print(f" predicted ratio = {s[0]/s[1]:.2f} : 1 : {s[2]/s[1]:.2f}")
print("measured 195 : 22 : 45 mb")
print(f" scaling by the first: 195/9 = {195/9:.1f} vs 22 measured ({abs(195/9-22)/22*100:.1f} % off)")
print(f" 195*2/9 = {195*2/9:.1f} vs 45 measured ({abs(195*2/9-45)/45*100:.1f} % off)")
print(f" and the parameter-free equality sigma(pi+p) = sigma(pi-n): "
f"{'exact' if abs(s[0]-s[3]) < 1e-12 else 'FAILS'}")decomposition from lib/cg.ts (Condon-Shortley), |pi N> in the |I,Iz> basis
|pi+ p > = +1.0000 |3/2,+3/2>
|pi- p > = +0.5774 |3/2,-1/2> -0.8165 |1/2,-1/2>
|pi0 n > = +0.8165 |3/2,-1/2> +0.5774 |1/2,-1/2>
|pi- n > = +1.0000 |3/2,-3/2>
cross-sections with A_1/2 = 0 (the Delta dominates):
pi+p -> pi+p |amp|^2 = 1.00000 (3.42)
pi-p -> pi-p |amp|^2 = 0.11111 (3.44)
pi-p -> pi0n |amp|^2 = 0.22222 (3.43)
pi-n -> pi-n |amp|^2 = 1.00000 (3.45)
predicted ratio = 9.00 : 1 : 2.00
measured 195 : 22 : 45 mb
scaling by the first: 195/9 = 21.7 vs 22 measured (1.5 % off)
195*2/9 = 43.3 vs 45 measured (3.7 % off)
and the parameter-free equality sigma(pi+p) = sigma(pi-n): exact Within 2 % and 4 %. For a prediction that used no dynamics — no potential, no coupling constant, no matrix element, nothing but how the states decompose — that is remarkable, and the residual is honest: at the peak is small but not zero, and the cross-sections carry backgrounds from the non-resonant part.
Turn it around and it becomes a measurement: the fact that the ratios come out 9 : 1 : 2 is how the isospin of the Δ was determined to be 3/2 in the first place. The pattern of the cross-sections identifies the resonance’s quantum number.
🔬 Experiment card — Fermi and collaborators, Chicago 1952
Apparatus
Pion beams of a few hundred MeV on a hydrogen target, with the cross-sections of and measured channel by channel as the beam energy is scanned. §4.2 returns to the apparatus and to the resonance itself; what matters here is the pattern in the numbers.What is measured
Three cross-sections at the same energy — elastic , elastic and the charge-exchange — and their variation with .The result
A large resonance at MeV in all of them, with peak cross-sections in the ratio 195 : 22 : 45 mb.What it proved
That the resonance has . The ratio 9 : 1 : 2 is what a pure amplitude predicts and what an one does not, so the isospin is read off the pattern of three numbers rather than measured directly — the same style of argument as everything else in this chapter.3.10 G-parity
C is a good quantum number only for completely neutral states (§3.3), which leaves the charged pions out: moves you to a different particle. G-parity G-parity G ≡ exp(−iπI_y)·C, a multiplicative quantum number for non-strange states of zero baryon number; G(π) = −1, so an n-pion state has G = (−1)ⁿ. Only the strong interaction conserves it. defined in §3.8-3.10 — open in glossary repairs that, by composing C with the isospin rotation that moves you back.
Bettini p. 125. C first, then a 180° rotation about the y-axis of isotopic space.
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Hover or tap a symbol above — it lights up in the equation and its meaning, units and type appear here.
🪜 Why G|π⟩ = −|π⟩ for all three charge states
Step 1 of 5 — Write the charge states as a vector(3.49)
Why you may do this: An I = 1 multiplet transforms like a vector in the three-dimensional isospin space, so it has Cartesian components π_x, π_y, π_z. The charge states are the spherical combinations of them — the same change of basis as between Cartesian and spherical unit vectors in ordinary space.
The book writes |π⁺⟩ with a leading PLUS. The standard Condon–Shortley convention (the one Appendix 4 and lib/cg.ts use) carries a leading minus. It changes nothing here — G comes out −1 either way, and every cross-section uses |amplitude|² — but the two conventions must not be mixed inside one derivation.
Bettini p. 125. The pions are the three Cartesian components of a vector in isospin space, and the calculation is what a 180° rotation does to a vector.
💡 What this really says — G-parity is a repair, and the repair matters more than the number
G-parity is a repair, and the pattern of the repair is more important than the quantum number.
C fails on the charged pions for one reason: it changes the charge, and therefore moves you along the multiplet. Isospin rotations also move you along the multiplet. So compose the two, choosing the rotation that exactly undoes C’s displacement, and the combination maps each state back to itself.
The cost is that G inherits the weaknesses of both its parents: only the strong interaction conserves G, because the electromagnetic interaction violates isospin and the weak interaction violates both isospin and C. A repaired symmetry is never better than its worst ingredient.
The practical use is the counting rule : a strong decay cannot change a three-pion state into a two-pion state. That is why the meson ω m = 782.66 MeV · Q = 0 · JP = 1− content uū, dd̄ τ / Γ = 8.68 ± 0.13 MeV open in the particle explorer (G = −1) decays to three pions and the meson ρ m = 775.26 MeV · Q = +1 · JP = 1− content ud̄, uū, dd̄, dū τ / Γ = 149.14 ± 0.8 MeV open in the particle explorer (G = +1) to two, and it is the tool §4.5 uses on the whole meson table.
Erratum — a mislabelled ket in the p. 125 derivation table
In the middle column of the derivation, the third row is labelled . That row starts from in the left-hand column and applies C, and — which is exactly the combination written. So the label should be ; it is left over from the first column. Every subsequent step and the result are correct.
A second, smaller wobble in the same table: (3.48) defines , while the worked lines apply . For integer isospin the two rotations are the same operator, so nothing downstream changes — but a reader checking the algebra will stop there.
🔑 If you remember only three things
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A multiplet is a list of states one interaction cannot distinguish. Three different nuclei with the same spin and parity at nearly the same energy are one object as far as the strong force is concerned.
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A broken symmetry can be repaired by combining it with another. G-parity exists because C fails on charged pions, and the repair is worth more than the quantum number it produces.
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Isospin borrows spin’s algebra and none of its meaning. Nothing rotates in space, and the only reason the notation transfers is that the group is the same.
Where this goes next
- The problems — eight of the chapter’s 31 exercises are isospin-ratio calculations, and the widget above answers them by construction.
- §4.2 The 3/2⁺ baryons is the Δ(1232) itself: the resonance whose dominance made 9 : 1 : 2 possible, treated as a particle rather than as a bump.
- §4.6–4.7 The quark model replaces SU(2) with SU(3), and with — the same block-diagonalisation, one dimension larger, and the reason the book insisted on the dimension notation here.
- §4.5 Pseudoscalar and vector mesons uses on every meson in the table.
- §7.11 The quark mixing matrix is what happens to flavour when the weak interaction — which respects none of this — takes over.
✅ Check yourself — isospin, cross-section ratios and G-parity
0/5 answered · 0 correct
1.Why does the electromagnetic interaction conserve but not ?
2.In the multiplet diagram above, press the antiparticle button (Question 3.3). What happens to the nucleon doublet?
Hint: flips every additive charge, and is built from two of them.
3.Isospin predicts at every energy. Why is this prediction stronger than the 9 : 1 : 2 result?
4.Set the widget's amplitude ratio to 0 and read the bars. The prediction gives 21.7 and 43.3 mb where the measurement gives 22 and 45. What should you conclude?
5.G-parity is defined as followed by a 180° rotation in isospin space. What does that composition buy, and what does it cost?